Given a number, the task is to check if number is divisible by 5. The input number may be large and it may not be possible to store even if we use long long int.
Input : n = 56945255 Output : Yes Input : n = 1234567589333150 Output : Yes Input : n = 3635883959606670431112222 Output : No
Since input number may be very large, we cannot use n % 5 to check if a number is divisible by 5 or not, especially in languages like C/C++. The idea is based on following fact.
A number is divisible by 5 if its digits last digit will be 0 or 5 .
For example, let us consider 769555 Number formed by last digit is = 5 Since 5 is divisible by 5 , answer is YES.
How does this work?
Let us consider 5335, we can write it as 5335 = 5*1000 + 3*100 + 3*10 + 5 The proof is based on below observation: Remainder of 10i divided by 5 is 0 if i greater than or equal to one. Note than 10, 100, 1000, ... etc lead to remainder 0 when divided by 5. So remainder of " 5*1000 + 3*100 + 3*10 + 5" divided by 5 is equivalent to remainder of following : 0 + 0 + 0 + 5 = 5 Since 5 is divisible by 5, answer is yes.
Below is the implementation of above idea.
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